用分配方法证明:(1)a^2-a+1的值恒为正；（2）-9X^2+8X-2的值恒小于0
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解决时间 2021-12-01 16:16
- 提问者网友：战皆罪
- 2021-11-30 21:29
用分配方法证明:(1)a^2-a+1的值恒为正；（2）-9X^2+8X-2的值恒小于0
最佳答案
- 五星知识达人网友：十年萤火照君眠
- 2021-11-30 21:46
a^2-a+1
=a^2-2*a*(1/2)+(1/2)^2-(1/2)^2+1
=(a-1/2)^2+3/4>=3/4>0
-9X^2+8X-2
=-9(x^2-8x/9)-2
=-9[x^2-2*x*(4/9)+(4/9)^2-(4/9)^2]-2
=-9(x-4/9)^2+9*16/81-2
=-9(x-4/9)^2-2/9<=-2/9<0
=a^2-2*a*(1/2)+(1/2)^2-(1/2)^2+1
=(a-1/2)^2+3/4>=3/4>0
-9X^2+8X-2
=-9(x^2-8x/9)-2
=-9[x^2-2*x*(4/9)+(4/9)^2-(4/9)^2]-2
=-9(x-4/9)^2+9*16/81-2
=-9(x-4/9)^2-2/9<=-2/9<0
全部回答
- 1楼网友：想偏头吻你
- 2021-12-01 00:07
1)a^2-a+1=(a-1/2)^2+3/4>0
2-9X^2+8X-2=-9(x-4/9)^2-2/9<0
2-9X^2+8X-2=-9(x-4/9)^2-2/9<0
- 2楼网友：孤独的牧羊人
- 2021-11-30 23:27
[a-(1/2)]^2+(3/4)>0
-9[x^2-(8/9x)+(2/9)]=-9{[x-(4/9)^2+(2/81)]
=-9[x-(4/9)]^2-(2/81)<0
-9[x^2-(8/9x)+(2/9)]=-9{[x-(4/9)^2+(2/81)]
=-9[x-(4/9)]^2-(2/81)<0
- 3楼网友：酒者煙囻
- 2021-11-30 23:09
a^2-a+1
=(a-1/2)^2+3/4
>=3/4
-9X^2+8X-2
=-9(x^2-8/9x+2/9)
=-9[(x-4/9)^2+2/81]
<=-2/9
=(a-1/2)^2+3/4
>=3/4
-9X^2+8X-2
=-9(x^2-8/9x+2/9)
=-9[(x-4/9)^2+2/81]
<=-2/9
- 4楼网友：北城痞子
- 2021-11-30 22:41
(1)a^2-a+1
=a^2-a+(1/2)^2-(1/2)^2+1
=(a-1/2)^2+3/4
>=3/4
>0
恒为正
(2)
-9X^2+8X-2
=-9(x^2-8/9x+2/9)
=-9(x^2-8/9x+16/81)+9*16/81-2
=-9(x-4/9)^2-18/81
<=-18/81=-2/9
<0
恒小于0
=a^2-a+(1/2)^2-(1/2)^2+1
=(a-1/2)^2+3/4
>=3/4
>0
恒为正
(2)
-9X^2+8X-2
=-9(x^2-8/9x+2/9)
=-9(x^2-8/9x+16/81)+9*16/81-2
=-9(x-4/9)^2-18/81
<=-18/81=-2/9
<0
恒小于0
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